Question: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{7q^2 - 21q - 28}{7q^2 - 63q - 70}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ a = \dfrac {7(q^2 - 3q - 4)} {7(q^2 - 9q - 10)} $ $ a = \dfrac{7}{7} \cdot \dfrac{q^2 - 3q - 4}{q^2 - 9q - 10} $ Simplify: $ a = \dfrac{q^2 - 3q - 4}{q^2 - 9q - 10}$ Next factor the numerator and denominator. $ a = \dfrac{(q + 1)(q - 4)}{(q + 1)(q - 10)}$ Assuming $q \neq -1$ , we can cancel the $q + 1$ $ a = \dfrac{q - 4}{q - 10}$ Therefore: $ a = \dfrac{ q - 4 }{ q - 10 }$, $q \neq -1$